Tìm x:a/ \(x^3=8\frac{27}{64}\)b/ \(2^x=8\frac{16}{32}\)c/ \(2^{2x-1}:4=8^3\)d/ \(x^{17}=x\)e/ \(\left(x-5\right)^4=\left(x-5\right)^6\)

TL

Trèo lên cột điện thế hiện tình yêu

5 tháng 11 2017

màu giải như cái lồnn vậy

EN

Emily Nain

22 tháng 12 2020

Tìm...

4. Tìm x biết: a) (3x - 7)5 = 0; b) \(\frac{1}{4}\) - (2x - 1)2 = 0; c) \(\frac{1}{16}\) - (5 - x)3 = \(\frac{31}{64}\); d) 2x = (3,8)3 : (-3,8)2; e) \(\left(\frac{27}{64}\right)^9\). x = \(\left(-\frac{3}{4}\right)^{32}\); f) \(5^{\left(x+5\right)\left(x^2-4\right)}\) = 1; g) (x - 2,5)2 = \(\frac{4}{9}\); h)...

#Toán lớp 8

1

TT

Thu Thao

22 tháng 12 2020

Rảnh rỗi thật sự .-.

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Đúng(2)

DB

Dương Bảo Thủy

7 tháng 8 2019

3

AH

Akai Haruma

Giáo viên

7 tháng 8 2019

a)

\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)

b)

\(\frac{1}{4}-(2x-1)^2=0\)

\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)

\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)

c)

\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)

\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)

\(\Leftrightarrow 5-x=\frac{-3}{4}\)

\(\Leftrightarrow x=\frac{23}{4}\)

d)

\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)

\(\Rightarrow x=3,8:2=1,9\)

AH

Akai Haruma

Giáo viên

7 tháng 8 2019

e)

\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)

\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)

\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)

\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)

f)

\(5^{(x+5)(x^2-4)}=1\)

\(\Leftrightarrow (x+5)(x^2-4)=0\)

\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)

g)

\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)

\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)

h)

\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)

\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)

BX

bui xuan dieu

19 tháng 3 2020

Tìm x, biết:

a) \(\left(\frac{-3}{4}\right)^{3x-1}=\frac{-27}{64}\)

b) \(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{265}\)

c) \(\frac{\left(x+3\right)^5}{\left(x+3\right)^2}=\frac{64}{27}\)

d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

1

S

santa

19 tháng 3 2020

a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)

\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)

\(\Leftrightarrow3x-1=3\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=\frac{4}{3}\)

b) Đề sai ! Sửa :

\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)

\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)

\(\Leftrightarrow2x+5=4\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\frac{1}{2}\)

c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)

\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)

\(\Leftrightarrow x+3=\frac{4}{3}\)

\(\Leftrightarrow x=-\frac{5}{3}\)

d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)

\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)

\(\Leftrightarrow x=\frac{4}{15}\)

AJ

agelina jolie

6 tháng 6 2016

tìm x biết...

#Toán lớp 6

3

PT

Phạm Tuấn Kiệt

6 tháng 6 2016

a) (x + 1/2) . (2/3 − 2x) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)

\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)

\(\Rightarrow x=-\frac{2}{11}\)

PT

Phạm Tuấn Kiệt

6 tháng 6 2016

c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)

\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)

\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)

\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)

\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)

\(\Rightarrow x=1\)

d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)

\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)

\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)

\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)

\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)

\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)

\(\Rightarrow x\approx28,7\) (số hơi lẻ)

e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)

1) Giải các pt sau: a) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\) b) \(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\) c) \(\frac{x+8}{6}-\frac{2x-5}{5}=\frac{x-1}{3}-x+7\) d) \(\frac{7x}{8}-5\left(x-9\right)=\frac{2x+1,5}{6}\) e) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\) f)...

VB

Vũ Bảo Duy

27 tháng 8 2020

a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)

b) \(\left(\frac{x}{3}-5\frac{1}{4}\right)^2-\frac{-2}{5}=1\frac{1}{25}\)

c) \(\frac{17-x}{12}=\frac{3}{17-x}\)

d) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)

#Toán lớp 6

0

AT

ánh tuyết nguyễn

8 tháng 2 2020

#Toán lớp 8

1

XU

Xích U Lan

8 tháng 2 2020

a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)

⇔ 3(x - 3) = 90 - 5(1 - 2x)

⇔ 3x - 9 = 90 - 5 + 10x

⇔ 3x - 10x = 90 - 5 + 9

⇔ -7x = 94

⇔ x = \(\frac{-94}{7}\)

S = { \(\frac{-94}{7}\) }

b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)

⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)

⇔ 6x - 4 - 60 = 9 - 6x - 42

⇔ 6x + 6x = 9 - 42 + 60 + 4

⇔ 12x = 31

⇔ x = \(\frac{31}{12}\)

S = { \(\frac{31}{12}\) }

c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7

⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210

⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210

⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40

⇔ 13x = 150

⇔ x = \(\frac{150}{13}\)

S = { \(\frac{150}{13}\) }

d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)

⇔ 21x - 120(x - 9) = 4(2x + 1,5)

⇔ 21x - 120x + 1080 = 8x + 6

⇔ 21x - 120x - 8x = 6 - 1080

⇔ -107x = -1074

⇔ x = \(\frac{1074}{107}\)

S = { \(\frac{1074}{107}\) }

e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5

⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840

⇔ 140x -140+56 -294x+42= 96x+48 -840

⇔ 140x -294x -96x = 48 -840 -42 -56+140

⇔ -250x = -750

⇔ x = 3

S = { 3 }

f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)

⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x

⇔ 4x+4+18x+9 = 4x+6x+6+7+12x

⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4

⇔ 0x = 0

S = R

Chúc bạn học tốt !

VN

Võ Ngọc Huyền

22 tháng 4 2020

Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html

Mình cảm ơn trước nhaa

Tìm x:a, \(\left(\frac{x}{3}-\frac{2}{5}\right):\frac{-2}{5}+\frac{1}{2}x=\frac{-3}{4}\)b. \(\left(\frac{-1}{8}x-\frac{3}{4}\right)-\frac{-8}{5}=\frac{5}{3}-x\)c. \(\left(x-\frac{1}{3}\right)^{x+1}=\left(x-\frac{1}{3}\right)^x\)d. \(\frac{x+2}{3}+\frac{2x-1}{4}\)e. \(\left(\frac{3}{2}-x\right)^4=64^2\)f. \(\left(5-\frac{x}{2}\right)^3-\frac{1}{27}=0\)Nhờ các bạn giúp mik tí...

ND

Ngô Đông Quỳnh

25 tháng 6 2019

0

NT

Nguyễn Trần Linh Na

21 tháng 7 2019

Tìm số nguyên x, nếu biết

\(\frac{^{2^{4-x}}}{16^5}=32^6\)

\(\frac{3^{2x+3}}{9^3}=9^{14}\)

\(\left(-2\right)^x=-\frac{\left(-8^4\right)}{\left(-32\right)^3}\)

\(\left(-5^x\right)=\frac{25^{10}}{\left(-5\right)^{17}}\)

1

EC

Edogawa Conan

21 tháng 7 2019

\(\frac{2^{4-x}}{16^5}=32^6\)

=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)

=> \(2^{4-x}=2^{30}.2^{20}\)

=> \(2^{4-x}=2^{50}\)

=> 4 - x = 50

=> x = 4 - 50 = -46

\(\frac{3^{2x+3}}{9^3}=9^{14}\)

=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)

=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)

=> \(3^{2x+3}=3^{28}.3^6\)

=> \(3^{2x+3}=3^{34}\)

=> 2x + 3 = 34

=> 2x = 34 - 3

=> 2x = 31

=> x = 31/2

NN

Ngọc Ngọc

5 tháng 3 2020

tìm x biết

a) \(\frac{x-1}{x+2}=\frac{4}{5}\left(x\ne-2\right)\) b)2^2x+1+4^x+3=264 c)\(\frac{x^2}{-8}=\frac{27}{x}\left(x\ne0\right)\) d)\(\frac{x+7}{-20}=\frac{-5}{x+7}\left(x\ne-7\right)\) e)\(\frac{x}{-8}=\frac{2}{-x^3}\left(x\ne0\right)\)

1

TC

Trên con đường thành công không có....

6 tháng 3 2020

a)Ta có:

\(\frac{x-1}{x+2}=\frac{4}{5}\Leftrightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Leftrightarrow5x-5=4x+8\)

\(\Leftrightarrow5x-4x=8+5\)

\(\Leftrightarrow x=13\)

b)Ta có:

\(2^{2x+1}+4^{x+3}=2^{2x+1}+2^{2x+6}=2^{2x+1}\left(1+2^5\right)=2^{2x+1}.33=264\Leftrightarrow2^{2x+1}=8=2^3\)\(\Rightarrow2x+1=3\Leftrightarrow2x=2\Leftrightarrow x=1\)

c)Ta có:

\(\frac{x^2}{-8}=\frac{27}{x}\Leftrightarrow x^3=-8.27=-216\Leftrightarrow x=-6\)

d)Ta có:

\(\frac{x+7}{-20}=\frac{-5}{x+7}\Leftrightarrow\left(x+7\right)^2=\left(-20\right)\left(-5\right)=100\Leftrightarrow\left[{}\begin{matrix}x+7=10\\x+7=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-17\end{matrix}\right.\)e)Ta có:

\(\frac{x}{-8}=\frac{2}{-x^3}\Leftrightarrow x.\left(-x^3\right)=-8.2\)

\(\Leftrightarrow-x^4=-16\Leftrightarrow x^4=16\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)